Tag: Trigonometria

0

Ficha de trabalho

9.º Ano: Trigonometria; Espaço - Outra Visão

A presente Ficha de Trabalho aborda os temas Trigonometria e Espaço – Outra Visão.

As dificuldades que encontres durante a sua resolução deves tentar superá-las consultando o manual e o caderno diário; depois, poderás tirar as dúvidas na aula ou na sala de estudo.

A realização da Ficha de Trabalho de forma empenhada contribuirá para uma preparação adequada para o Teste de Avaliação.…

Duas funções trigonométricas 0

Duas funções trigonométricas

Enunciado

Na figura estão as representações gráficas de duas funções, f e g, no intervalo $\left[ -2\pi ,2\pi  \right]$.

Sabe-se que:

  • f é definida por $f(x)=sen\,x$;  
  • g é definida por $g(x)=\cos (3x)$;  
  • A é um ponto de intersecção dos gráficos de f e de g.

Sem recorrer à calculadora, a não ser para efectuar eventuais cálculos numéricos, determine a abcissa (valor exacto) de A.…

Duas condições trigonométricas 0

Duas condições trigonométricas

Proposta 34 - Utilizando as capacidades da calculadora gráfica

Enunciado Resolva, utilizando as capacidades da sua calculadora gráfica, as seguintes condições:

  1. $sen\,x=0,4$
     
  2. $sen\,x>0,3$

Resolução >> Resolução

  1. Consideremos duas funções f1 e f2, reais de variável real, de domínio R, definidas por ${{f}_{1}}(x)=sen\,x$ e ${{f}_{2}}(x)=0,4$.
     
    Efectuada a representação gráfica destas funções no intervalo $\left[ -2\pi ,2\pi  \right]$, verificamos que, no intervalo $\left[ 0,2\pi  \right]$, os gráficos intersectam-se nos pontos A e B, de coordenadas indicadas na figura (a abcissa é um valor aproximado às milésimas).
Funções trigonométricas 1

Funções trigonométricas

var parameters = { "id": "ggbApplet", "width":970, "height":510, "showMenuBar":false, "showAlgebraInput":false, "showToolBar":false, "customToolBar":"0 39 | 1 501 67 , 5 19 , 72 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 | 16 51 64 , 70 | 10 34 53 11 , 24 20 22 , 21 23 | 55 56 57 , 12 | 36 46 , 38 49 50 , 71 | 30 29 54 32 31 33 | 17 26 62 73 , 14 68 | 25 52 60 61 | 40 41 42 , 27 28 35 , 6", "showToolBarHelp":false, "showResetIcon":true, "enableLabelDrags":false, "enableShiftDragZoom":false, "enableRightClick":false, "errorDialogsActive":false, "useBrowserForJS":false, "preventFocus":false, "language":"pt", // use this instead of ggbBase64 to load a material from GeoGebraTube // "material_id":12345, "ggbBase64":"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"}; // is3D=is 3D applet using 3D view, AV=Algebra View, SV=Spreadsheet View, CV=CAS View, EV2=Graphics View 2, CP=Construction Protocol, PC=Probability Calculator, DA=Data Analysis, FI=Function Inspector, PV=Python, macro=Macro View var views = {'is3D': 0,'AV': 0,'SV': 0,'CV': 0,'EV2': 0,'CP': 0,'PC': 0,'DA': 0,'FI': 0,'PV': 0,'macro': 0}; var applet = new GGBApplet(parameters, '5.0', views); window.onload = function() {applet.inject('ggbApplet')};

 …

0

Um polígono [ABEG]

Trigonometria: Infinito 11 A - Parte 1 Pág. 99 Ex. 72

Enunciado

Na figura está representado, a cor, um polígono [ABEG].
Tem-se que:

  • [ABFG] é um quadrado de lado 2.  
  • FD é um arco de circunferência de centro em B; o ponto E move-se ao longo desse arco; em consequência, o ponto C desloca-se sobre o segmento [BD], de tal forma que se tem sempre $[EC]\bot [BD]$.  
Equações trigonométricas 5 0

Equações trigonométricas 5

Trigonometria: Infinito 11 A - Parte 1 Pág. 99 Ex. 71

Enunciado

Resolva as seguintes equações trigonométricas, no intervalo indicado:

  1. $-\sqrt{3}-2\,sen\,\theta =0$ para $\theta \in \left[ -\pi ,\pi  \right]$
     
  2. $-2+\sqrt{3}\,tg\,\theta =1$ para $\theta \in \left[ 0,2\pi  \right]$
     
  3. $1+\sqrt{2}\cos \theta =3$ para $\theta \in \left[ \pi ,3\pi  \right]$
     
  4. $4{{\cos }^{2}}\theta =3$ para $\theta \in \left[ -\pi ,\pi  \right]$

Resolução >> Resolução

  1. Ora, \[\begin{array}{*{35}{l}}    -\sqrt{3}-2\,sen\,\theta =0 & \Leftrightarrow  & sen\,\theta =-\frac{\sqrt{3}}{2}  \\    {} & \Leftrightarrow  & \begin{array}{*{35}{l}}    \theta =-\frac{\pi }{3}+2k\pi  & \vee  & \theta =(\pi -(-\frac{\pi }{3}))+2k\pi \,,\,\,k\in \mathbb{Z}  \\ \end{array}  \\    {} & \Leftrightarrow  & \begin{array}{*{35}{l}}    \theta =-\frac{\pi }{3}+2k\pi  & \vee  & \theta =\frac{4\pi }{3}+2k\pi \,,\,\,k\in \mathbb{Z}  \\ \end{array}  \\ \end{array}\] donde, para os valores indicados de k, se obtém: \[\begin{matrix}    k=-1: & \theta =-\frac{7\pi }{3} & \vee  & \theta =-\frac{2\pi }{3}  \\    k=0: & \theta =-\frac{\pi }{3} & \vee  & \theta =\frac{4\pi }{3}  \\    k=1: & \theta =\frac{5\pi }{3} & \vee  & \theta =\frac{10\pi }{3}  \\ \end{matrix}\] Logo, no intervalo considerado, o conjunto-solução da equação é $S=\left\{ -\frac{2\pi }{3},-\frac{\pi }{3} \right\}$.
Mostre que 0

Mostre que

Trigonometria: Infinito 11 A - Parte 1 Pág. 99 Ex. 70

Enunciado

Mostre que: $(\cos \alpha -sen\,\alpha )(\cos \alpha +sen\,\alpha )-1=-2\,se{{n}^{2}}\,\alpha $.

Resolução >> Resolução

\[\begin{array}{*{35}{l}}
   (\cos \alpha -sen\,\alpha )(\cos \alpha +sen\,\alpha )-1 & = & {{\cos }^{2}}\alpha +\cos \alpha \times sen\,\alpha -\cos \alpha \times sen\,\alpha -se{{n}^{2}}\,\alpha -1  \\
   {} & = & {{\cos }^{2}}\alpha -se{{n}^{2}}\,\alpha -1  \\
   {} & = & -(1-{{\cos }^{2}}\alpha )-se{{n}^{2}}\,\alpha   \\
   {} & = & -se{{n}^{2}}\,\alpha -se{{n}^{2}}\,\alpha   \\
   {} & = & -2\,se{{n}^{2}}\,\alpha   \\
\end{array}\]

<< Enunciado
Calcule o valor exacto da expressão 2

Calcule o valor exacto da expressão

Trigonometria: Infinito 11 A - Parte 1 Pág. 99 Ex. 69

Enunciado

Calcule o valor exacto da expressão: $sen\,\frac{13\pi }{4}+\cos 5\pi -tg\,(-7\pi )+\cos (-\frac{23\pi }{4})$.

Resolução >> Resolução

\[\begin{array}{*{35}{l}}
   sen\,\frac{13\pi }{4}+\cos 5\pi -tg\,(-7\pi )+\cos (-\frac{23\pi }{4}) & = & sen\,(2\pi +\pi +\frac{\pi }{4})+\cos \pi -tg\,(0)+\cos (-6\pi +\frac{\pi }{4})  \\
   {} & = & -\,sen\,(\frac{\pi }{4})+\cos \pi -tg\,(0)+\cos (\frac{\pi }{4})  \\
   {} & = & -\frac{\sqrt{2}}{2}-1-0+\frac{\sqrt{2}}{2}  \\
   {} & = & -1  \\
\end{array}\]

 

<< Enunciado
Simplifique a expressão 0

Simplifique a expressão

Trigonometria: Infinito 11 A - Parte 1 Pág. 99 Ex. 68

Enunciado

Simplifique a expressão:

$-2\,sen\,(\alpha +\frac{\pi }{2})+\cos (\frac{5\pi }{2}-\alpha )-sen\,(-\alpha +\pi )$

Resolução >> Resolução

\[\begin{array}{*{35}{l}}
   -2\,sen\,(\alpha +\frac{\pi }{2})+\cos (\frac{5\pi }{2}-\alpha )-sen\,(-\alpha +\pi ) & = & -2\,sen\,(\frac{\pi }{2}-(-\alpha ))+\cos (2\pi +\frac{\pi }{2}-\alpha )-sen\,(\pi -\alpha )  \\
   {} & = & -2\cos (-\alpha )+sen\,\alpha -sen\,\alpha   \\
   {} & = & -2\cos \alpha   \\
\end{array}\]

<< Enunciado
Sabe-se que… 0

Sabe-se que…

Trigonometria: Infinito 11 A - Parte 1 Pág. 99 Ex. 67

Enunciado

Sabe-se que $\cos \alpha =\frac{1}{3}$ e que $\frac{3\pi }{2}<\alpha <2\pi $.

Determine o valor exacto de:

  1. $sen\,\alpha $
     
  2. $tg\,(\pi -\alpha )$

Resolução >> Resolução

  1. Tendo em consideração a FFT e que $\alpha \in 4.{}^\text{o}Q$, temos \[sen\,\alpha =-\sqrt{1-{{(\frac{1}{3})}^{2}}}=-\sqrt{\frac{9-1}{9}}=-\frac{2\sqrt{2}}{3}\]
     
  2. Ora, \[tg\,(\pi -\alpha )=tg\,(-\alpha )=-tg\,\alpha =\frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}}=-2\sqrt{2}\]
<< Enunciado